基于相似产品先验的产品Bayes可靠性验证

doi:10.12305/j.issn.1001-506X.2026.06.18

系统工程与电子技术 ›› 2026, Vol. 48 ›› Issue (6): 1972-1979.doi: 10.12305/j.issn.1001-506X.2026.06.18

基于相似产品先验的产品Bayes可靠性验证

王昱¹, 邱浩波¹^,*, 何小二², 王铭靖¹, 蒋琛¹, 高亮³

1. 华中科技大学机械科学与工程学院，湖北武汉 430074
2. 武汉第二船舶设计研究所，湖北武汉 430205
3. 华中科技大学国家智能设计与数控创新中心，湖北武汉 430074

收稿日期:2025-02-21 修回日期:2025-05-09 出版日期:2026-06-25 发布日期:2026-03-19
通讯作者: 邱浩波
作者简介:王　昱（2001—），女，硕士研究生，主要研究方向为复杂装备可靠性验证
何小二（1989—），男，高级工程师，硕士，主要研究方向为舰船总体设计
王铭靖（1999—），男，硕士研究生，主要研究方向为复杂装备安全性分析
蒋　琛（1993—），男，讲师，博士，主要研究方向为装备不确定性设计与决策调控
高　亮（1974—），男，教授，博士，主要研究方向为智能优化算法及其在设计与制造中的应用
基金资助:
国家自然科学基金（52405268，52475260）资助课题

Bayes reliability demonstration of products based on priors from similar products

Yu WANG¹, Haobo QIU¹^,*, Xiaoer HE², Mingjing WANG¹, Chen JIANG¹, Liang GAO³

1. State Key Laboratory of Intelligent Manufacturing Equipment and Technology，Huazhong University of Science and Technology，Wuhan 430074，China
2. Wuhan Second Ship Design and Research Institute，Wuhan 430205，China
3. National Center of Technology Innovation for Intelligent Design and Numerical Control，Huazhong University of Science and Technology，Wuhan 430074，China

Received:2025-02-21 Revised:2025-05-09 Online:2026-06-25 Published:2026-03-19
Contact: Haobo QIU

摘要/Abstract

摘要：

为了解决长寿命、高可靠性产品在可靠性验证试验中面临的耗时长问题，本文以寿命服从指数分布的产品为研究对象，提出了一种基于相似产品先验的Bayes可靠性验证试验方案设计方法。首先，引入贝叶斯方法，将同类型产品的试验数据作为先验信息，采用拟合优度检验统计量量化产品之间的相似性，进一步利用继承因子实现先验分布的计算。随后，运用贝叶斯平均风险准则计算实际的生产方和使用方风险。最后，通过迭代方法求解最优的可靠性验证试验方案。实验结果表明，与GJB899A-2009标准中的可靠性验证试验方案相比，本文方法可显著缩短试验时间。

关键词: 可靠性验证试验, Bayes方法, 相似产品, 继承因子

Abstract:

To address the prolonged duration associated with reliability demonstration tests for long-life and high-reliability products, this paper focuses on products whose lifetimes follow exponential distributions and proposes a Bayes reliability demonstration test （RDT） design method incorporating prior information obtained from similar products. Firstly, the Bayesian method is introduced into the RDT design. Concretely, the experimental data achieved from similar products is viewed as prior information. The similarity between any two products is quantified by goodness-of-fit test statistics, and prior distribution is calculated through an inheritance factor. Secondly, the Bayes average risk criterion is utilized to describe the risks for both the manufacturer and the consumer. Finally, an optimal RDT plan is obtained through a numerical iteration method. Compared with the RDT plan in the GJB899A-2009 standard, the experimental results exhibit that the devised method can significantly reduce the test time.

Key words: reliability demonstration test, Bayes methods, similar products, inheritance factor

中图分类号:

TB 114

王昱, 邱浩波, 何小二, 王铭靖, 蒋琛, 高亮. 基于相似产品先验的产品Bayes可靠性验证[J]. 系统工程与电子技术, 2026, 48(6): 1972-1979.

Yu WANG, Haobo QIU, Xiaoer HE, Mingjing WANG, Chen JIANG, Liang GAO. Bayes reliability demonstration of products based on priors from similar products[J]. Systems Engineering and Electronics, 2026, 48(6): 1972-1979.

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产品编号	失效间隔时间/h
A	13, 17, 23, 10, 39, 61, 16, 97, 144, 54, 100, 157, 22, 162, 46
B	83, 141, 35, 224, 88

α₀	β₀	T/h	C	α	β
0.1	0.1	761	10	0.090	0.099
0.1	0.1	820	11	0.075	0.099
0.1	0.1	879	12	0.063	0.099
0.1	0.1	937	13	0.043	0.099

α₀	β₀	T/h	C	α	β
0.2	0.2	329	4	0.176	0.199
0.2	0.2	388	5	0.140	0.198
0.2	0.2	445	6	0.111	0.199
0.2	0.2	503	7	0.090	0.198

α₀	β₀	T/h	C	α	β
0.3	0.3	119	1	0.286	0.297
0.3	0.3	176	2	0.209	0.299
0.3	0.3	232	3	0.157	0.299
0.3	0.3	288	4	0.120	0.297

T/h	C	$ {\alpha }_{0} $	$ {\beta }_{0} $	$ {\alpha }^{\prime} $	$ {\beta }^{\prime} $	$ \alpha $	$ \beta $
945	13	0.1	0.1	0.098	0.102	0.055	0.093
624	9	0.1	0.2	0.102	0.201	0.062	0.186
394	5	0.2	0.2	0.207	0.200	0.147	0.187
185	2	0.3	0.3	0.282	0.285	0.230	0.268

基于相似产品先验的产品Bayes可靠性验证

Bayes reliability demonstration of products based on priors from similar products

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摘要/Abstract

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参考文献 31

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双方风险	899A	本文方法	改进效果/%
$ {\alpha }_{0}=0.1,{\beta }_{0}=0.1 $	（945,13）	（761,10）	19.470
$ {\alpha }_{0}=0.1,{\beta }_{0}=0.2 $	（624,9）	（503,7）	19.391
$ {\alpha }_{0}=0.2,{\beta }_{0}=0.2 $	（394,5）	（329,4）	16.497
$ {\alpha }_{0}=0.3,{\beta }_{0}=0.3 $	（185,2）	（119,1）	35.675

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